For what emf $\epsilon$does the $200\Omega$resistor in$figureP28.64$ dissipate no power? Should the emf be oriented with its positive terminal at the top or at the bottom?

We should find oriented with its positive terminal at the top or at the bottom.

When the current flows through a resistor, it dissipates energy, the rater of the dissipated energy is the power. This rate where the energy is transferred to the current to the resistor is

$P_R=I∆V_R$

The dissipated power is zero when the current through the resistor $200\Omega$is zero (no current flow). But at the condition, the current will not flow through the $200\Omega$? This occurs when the right battery $\epsilon$applies a current through $300\Omega$with the same magnitude of the left battery but in opposite direction. So, let us find the current that is flow due the left battery $50V$using Ohm's law.

Ohm's law states that the current Iflows through a resistance due to the potential difference V between resistance.

$I=\frac{V}{R}$

So, we get the current through $100\Omega$by using the values $50Vand100\Omega$into equation $\left(2\right)$

$I_{100\Omega }=\frac{V}{R}=\frac{50V}{100\Omega }=0.5A$

As we mentioned above, this is the current that flows in $300\Omega$to make no current flows through $200\Omega$. So, the voltage that exert this current through the resistor $300\Omega$is calculated by Ohm's law in the form

$$\begin{gathered} \epsilon =IR \\ =(0.5A)\left(300\Omega \right) \\ =150V \end{gathered}$$

The right battery has emf $\epsilon =150V$and to exerts a current in the opposite direction, its terminal positive should be bottom.